physics

As shown in the diagram, a bullet of mass 7.0 g strikes a block of wood of mass 2.1 kg. The block of wood is suspended by a string of length 2.0 m, forming a pendulum. The bullet lodges in the wood, and together they swing upward a distance of 0.40 m from the original position of the wood block. What was the speed of the bullet just before it struck the wooden block?

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asked by jose
  1. bullet mass + wood mass = 2.107 kg

    potential energy if up 0.4 meter:

    I assume this is along the circumference
    so
    .4 = theta * 2 =
    theta = .2 radians = 11.5 degrees

    distance up from bottom = 2 (1-cos theta)
    = .0803 meters

    so energy at top = m g h =2.107*9.81*.0803 = 1.66 Joules

    that is its kinetic energy at the bottom
    (1/2) m v^2 = 1.66
    .5*2.107*v^2 = 1.66
    v = 1.26 m/s

    momentum of block with bullet
    = 2.107 * 1.26 = 2.64 kg m/s

    that was the bullet momentum before the crash
    .007 v = 2.64
    v = 378 m/s

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    posted by Damon

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